The number of gallons of water in the tank at t=10 is
... W(10) = 160,000 -10(8000 -10) = 80100
The number of gallons of water in the tank at t=10.5 is
... W(10.5) = 160,000 -10.5(8000 -10.5) = 76110.25
The rate of change over the interval is
... (W(10.5) - W(10))/(10.5 - 10) = (76110.25 - 80100)/(0.5) = -7979.5
The average rate of change in the number of gallons of water in the tank over the interval is -7979.5 gal/min.
The sign is negative, so the amount of water is decreasing.
Answer:
The correct answer is 15 cm.
Step-by-step explanation:
Let the width of the required poster be a cm.
We need to have a 6 cm margin at the top and a 4 cm margin at the bottom. Thus total margin combining top and bottom is 10 cm.
Similarly total margin combining both the sides is (4+4=) 8 cm.
So the required printing area of the poster is given by {( a-10 ) × ( a - 8) }
This area is equal to 125 as per as the given problem.
∴ (a - 10) × (a - 8) = 125
⇒ - 18 a +80 -125 =0
⇒ - 18 a -45 = 0
⇒ (a-15) (a-3) = 0
By law of trichotomy the possible values of a are 15 and 3.
But a=3 is absurd as a 4.
Thus the required answer is 15 cm.